Problem: Rewrite the equation by completing the square. $4 x^{2} -12 x +9 = 0$ $(x + $
Solution: $\begin{aligned} 4 x^2 -12 x +9&=0 \\\\ 4 x^2 -12 x &=-9 \\\\ x^2 -3 x&=-\dfrac{9}{4} \end{aligned}$ Now we want to complete $x^2 -3 x$ into a perfect square. To do that, we should add $\left(\dfrac{{-3}}{2}\right)^2={\dfrac{9}{4}}$ to it: $x^2{-3}x + {\dfrac{9}{4}}=\left(x -\dfrac{3}{2} \right)^2$ $\begin{aligned} x^2 -3 x&=-\dfrac{9}{4} \\\\ x^2 -3 x + {\dfrac{9}{4}}&=-\dfrac{9}{4} + {\dfrac{9}{4}} \\\\ \left(x -\dfrac{3}{2} \right)^2&=0 \end{aligned}$ In conclusion, the equation after completing the square is written as: $\left(x -\dfrac{3}{2} \right)^2=0$